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$f \geq 0 \ on \ [a,b] \ RI.$

Then, $$\int_{a}^{b} f(x) dx \geq 0$$

I have seen this

Is the Riemann integral of a strictly positive function positive?

However, I'm not fully satisfied with the answers there as the conditions are slightly different. In the accepted answer, they say that since $f$ is continuous and positive on I (to make it similar to my problem let $I = [a,b]$, then $\int_{I} f > 0$. However, this is the conclusion. To me, it seems like they used the final answer to prove the result.

In my problem, however, I am not told $f$ is continuous in my problem. So that is a major assumption I am missing because while every continuous function is RI, being RI does not guarantee continuity.

I did look at a second answer on that same problem that has more upvotes. However, they prove it by assuming that $\int_{0}^{1} f = 0$. This works for that problem because $f > 0$ strictly. In my problem, however, it could be equal to $0$. So showing this case won't really help me here.

There was one more answer, but that uses Lebesgue integration, which I have not covered in this class.

I am really not so sure how to start it myself. I am convinced it's true after trying to come up with counterexamples because I wasn't convinced at first, but I really have no idea how to show it. Can anyone help me with this one? Thanks!

Nolan P
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    Start with the definition of a Riemann integral. If you plug in the assumption $f \ge 0$ into that definition, this result pops out basically immediately. – Trevor Gunn Dec 13 '20 at 15:14
  • Is it because the supremum and the infimum have to be positive since f is positive everywhere? That's the only thing I really see in the definition right now that could do it. – Nolan P Dec 13 '20 at 15:20
  • You have it right and you should be more confident in your answer. And if you're not confident then keep breaking down your proof step by step and checking it against the definitions/theorems you know. If $f \ge 0$ then $0 \le \inf f \le \sup f$. If you're not 100% sure why this holds, go back to your notes/textbook and check. Check that any lower sum is always less than or equal to the integral if you're unsure about that. – Trevor Gunn Dec 13 '20 at 15:51
  • I see. Thank you for your help. I will say my confidence has lowered throughout the semester, but that's for other reasons you don't need to be concerned with. Did you want to type this out as an answer so I can accept it? In case if others want to see it in the future. If not I think I may just delete the question since it turned out to be quite simple. – Nolan P Dec 13 '20 at 16:11

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The definition of the Riemann integral looks something like

$$ \sup_{\mathcal{P}} \sum_{i=1}^n \left(\inf_{[x_{i},x_{i+1}]} f(x) \right) (x_i - x_{i-1}) = \int_a^b f = \inf_{\mathcal{P}} \sum_{i=1}^n \left(\sup_{[x_{i},x_{i+1}]} f(x) \right) (x_i - x_{i-1}). $$

You might write this somewhat differently but the point is that every piece of this is $\ge 0$ by assumption. Specifically, if $f \ge 0$ then $\inf f \ge 0$ on any interval.

Trevor Gunn
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